I got this equation for matrix element, no summation notation, just free indices: $$ \varepsilon_{ij} = \dfrac{u_i}{k_j}+\dfrac{u_j}{k_i}$$

I want to sum all the elements of this matrix multiplied in some manner for which, I belive, I can use summation notation. From now on I want to use i,j as dummy indices: $$ \varepsilon_{ij} = \dfrac{u_i}{k_j}+\dfrac{u_j}{k_i} \quad \mid \cdot k_ik_j \\ k_ik_j \varepsilon_{ij} = \dfrac{k_i k_ju_i}{ k_j}+\dfrac{k_ik_ju_j}{ k_i} = k_iu_i + k_ju_j$$

I guess it is legitimate.

Why can't I then write $k_i u_i$ as a vectors dot product: $k_ik_j\varepsilon_{ij} = k_iu_i + k_ju_j = \vec k \cdot \vec u+ \vec k \cdot \vec u = 2\vec k \cdot \vec u$ and get back to my initial matrix?

$$k_ik_j \varepsilon_{ij} = 2\vec k \cdot \vec u \quad \mid : (k_ik_j) \\ \varepsilon_{ij} = \dfrac{2\vec k \cdot \vec u }{k_ik_j }$$ This is obviously not true, as:

$$\varepsilon_{xz} = \dfrac{2\vec k \cdot \vec u }{k_xk_z } = 2\dfrac{k_xu_x+k_yu_y+k_zu_z}{k_xk_z} \neq \dfrac{u_x}{k_z}+ \dfrac{u_z}{k_x}$$

  • 2
    $\begingroup$ Your last step is the problem- you can‘t divide by $k_ik_j$, because there is a sum over all $i$ and $j$ on the left side. $\endgroup$
    – Benjamin
    Sep 1 '20 at 15:53
  • $\begingroup$ Thank you! Please write an answer and I will accept it. $\endgroup$
    – Paweł J
    Sep 1 '20 at 16:38

The short answer: Dividing the equation $k_ik_j\varepsilon_{ij}=2\vec k\cdot\vec u$ by $k_ik_j$ is not allowed, because you free the summation indices $i$ and $j$.

A little longer answer: The matrix definition $\varepsilon_{ij} = \dfrac{u_i}{k_j}+\dfrac{u_j}{k_i}$ consists of $d^2$ equations, when $d$ is the dimension. After summing up you are at the single equation $k_ik_j\varepsilon_{ij}=2\vec k\cdot\vec u$. There is no way to conclude on $d^2$ equations from there, when $d>1$.

A matrix notation answer: In matrix language rather than Einstein notation the equation $k_ik_j\varepsilon_{ij}=2\vec k\cdot\vec u$ could be written as $$ \vec k^T\varepsilon\vec k = 2\vec k\cdot\vec u. $$ It is quite obvious out of this perspective, that there is no way to find all matrix elements of $\varepsilon$ from there, unless $\varepsilon$ is a $1\times1$ matrix.

  • $\begingroup$ Great answer, thank you! $\endgroup$
    – Paweł J
    Sep 1 '20 at 20:30

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